IB DIPLOMA PROGRAMME
M06/4/PHYSI/SP2/ENG/TZ1/XX/M+
PROGRAMME DU DIPL�ME DU BI
PROGRAMA DEL DIPLOMA DEL BI
c
MARKSCHEME
May 2006
PHYSICS
Standard Level
Paper 2
16 pages
  2  M06/4/PHYSI/SP2/ENG/TZ1/XX/M+
This markscheme is confidential and for the exclusive use of
examiners in this examination session.
It is the property of the International Baccalaureate and must not
be reproduced or distributed to any other person without the
authorization of IBCA.
  6  M06/4/PHYSI/SP2/ENG/TZ1/XX/M+
SECTION A
A1. (a) 17.0
16.0
15.0
14.0
13.0
h / cm
12.0
11.0
10.0
9.0
8.0
0 10 20 30 40 50 60 70 80 90
�
� / C
(i) sensible line of fit with reasonable distribution of points either side of line; [1]
(ii) h0 = 16.2(ą 0.2) cm ; [1]
(b) this is a straight-line graph so has equation of the form y = mx + c ;
h = h0(1- k� ) gives h = h0 - h0k� ;
m = -h0k ;
c = h0 ; [4]
essentially look for:
stating equation of a straight-line graph, showing that h = h0(1- k� ) can be written
in this form, identifying m and c.
  7  M06/4/PHYSI/SP2/ENG/TZ1/XX/M+
If credit for m = -h0k has not been given in (b)
gradient �#
(c) k = ;
Ź#then
h0 �# it can be given if this statement is correct.
gradient = 0.077 (ą 0.003) ;
0.077
therefore, k = ;
16.2
H" 4.8�10-3 deg C-1
Or
Allow use of a point on line of best fit i.e. choice of data point from line of best fit;
correct substitution into h = h0(1 - k�);
1 h
correct rearrangement essentially showing that k = (1- ) ; [3]
� h0
Accept range of answers for the gradient between 4.9 �10-3 and 4.6 �10-3 .
(d) estimate:
gradient
h = ;
r
gradient =1.5(ą 0.2)�10-5 ;
1.5�10-5
r == 6.0(ą0.8)�10-7 m ;
25
comment:
this is very small so it is unlikely that capillary action is the only mechanism /
OWTTE / this assumes that the direct proportion holds for values of h up to 25m /
OWTTE; [4]
Accept ECF based on estimate only if comment is reasonable and consistent. If numerical
value is correct, then award the mark for a plausible explanation (e.g. reference to molecular
forces) as to why this is a reasonable value.
  8  M06/4/PHYSI/SP2/ENG/TZ1/XX/M+
A2. (a) there are no positions;
the lamp is effectively in series with 100 k&! no matter what the position of S;
this means that the pd across it will always be close to zero (very small) / never
reach 6 V;
Or
there are no positions;
the resistance of the filament is much smaller that 100 k&! ;
so (nearly) all the potential of the battery appears across the variable resistance; [3]
Award [1] max for correct answer with no argument or incorrect argument.
Anthropomorphic answers such  battery has a lot of resistance to overcome
score [1] max. Must mention that voltmeter is effectively in series with battery to
get full marks.
V
(b) I = ;
R
12
= = 0.80 A ; [2]
15
(c)
S
12 V
A
V
correct position of ammeter;
correct position of voltmeter (either to the right or left of the lamp); [2]
A3. (a) towards the centre of Earth; [1]
(b) gravity/gravitational/centripetal; [1]
(c) T = 24�3600 = 8.64�104 s ;
4Ą2R 4Ą2 � 4.2�107
substitute into a = = ;
2
T (8.64)2 �108
= 0.23ms-2 ; [3]
  9  M06/4/PHYSI/SP2/ENG/TZ1/XX/M+
SECTION B
B1. (a) the amount of energy / heat required to raise the temperature of 1 kg of a substance
through 1 K;
Or
Amount of energy needed to increase the temperature of a unit mass by one
degree. [1]
(b) the internal energy is the total energy of the molecules of a substance;
the greater the specific heat (the more energy required to raise unit mass through
1 K this) means that to increase the temperature by the same amount, more energy
most be given to substance A than to substance B (so internal energy is greater) /
OWTTE; [2]
Award [0] for correct answer with incorrect or no explanation.
(c) (i) so that the metal reaches the temperature of the boiling water; [1]
(ii) QM = mass of metal�specific heat capacity of metal�fall in temperature of metal; [1]
(iii) QW = mass of water�specific heat capacity of water � rise in temp of water; [1]
(iv) because energy is lost to the surroundings; [1]
(d) (i) (energy is transferred) by conduction through the insulation of the element /
OWTTE;
(energy is then transferred) by the bulk motion of the water / convection
through the water / OWTTE;
the element will also radiate some energy which will be absorbed by the
water / OWTTE; [3]
(ii) energy supplied by heater in 1s = 7.2�103 J ;
energy per second = mass per second � sp ht � rise in temperature;
7.2�103 = mass per second � 4.2�103 � 26 ;
to give mass per second = 0.066 kg / flow rate = 0.066kg s-1; [4]
(iii) energy is lost to the surroundings;
flow rate is not uniform; [2]
Do not allow  the heating element is not in contact with all the water flowing in the unit .
Accept answers that imply that there will be a temperature gradient between element
and wall of pipe. Do not accept answers such as  element will not heat water
uniformly .
P
(iv) P = VI, I = ;
V
7.2�103
== 30 A ; [2]
240
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(v) when operating at 7.2 kW the element is at a higher temperature / hotter
than when first switched on;
therefore, resistance is greater (and so current is smaller) / OWTTE;
Or
element is cold / OWTTE when first switched on;
therefore, smaller resistance than when hot (and so current is larger); [2]
2
V
(e) (i) P = ;
R
2402 1102
= ;
R240 R110
2
R110 110
�# ś#
= ;
ź#
R240 ś# 240
�# #
= 0.21
Or
from P = VI
11
240I2 = 110I1 to give I2 = I1;
24
I22 R2 = I12R1;
2
R1 I22 11
�# ś#
= = ;
ź#
R2 I12 ś# 24
�# #
= 0.21 [3]
(ii) to get equivalent power, heating elements must have lower resistance;
therefore, they have to be physically larger so more expensive / take up more
space;
Or
smaller voltage supply needs larger current;
so thicker cables therefore, more expensive / take up more space; [2]
  11  M06/4/PHYSI/SP2/ENG/TZ1/XX/M+
B2. Part 1 Momentum
(a) if the total (or net) external force acting on a system is zero / for an isolated system;
the momentum of the system is constant/momentum before collision equals
momentum after collision; [2]
Award [1] for  momentum before (collision) = momentum after (collision) .
(b) (i) (a collision in which) kinetic energy is not lost / kinetic energy is conserved;
[1]
(ii) the momentum of the puck is not conserved since a force acts on it during
collision / OWTTE;
the rink is attached to the Earth and momentum is given to the Earth such
that the change in momentum of the puck is equal to the change in
momentum of the Earth / OWTTE;
Or
the momentum of the Earth and puck are conserved / OWTTE;
the change in momentum of the puck is equal and opposite to the change in
momentum of the Earth; [2]
This is a discussion so more than bald statements are required e.g. identification
of system and some explanation.
(c)
vector 5.0 cm long;
at right angles to initial vector as shown;
By eye is sufficient.
resultant vector as shown;
stated length = 7.1(ą 0.2) cm equivalent to 0.71(ą 0.2) Ns ;
Length should be checked.
  12  M06/4/PHYSI/SP2/ENG/TZ1/XX/M+
Or
Second vector at right angles to first;
And of equal length;
Difference shown as a vertical vector;
Of magnitude 0.52 + 0.52 ;
= 0.71 N s [4]
Caution: Many students are obtaining instead the sum of the two momenta rather than the
difference. In this case the numerical answer is the same for the magnitude so watch out.
"p 0.71
(d) F = = = 59 N ;
"t 12�10-3
this is the average force and from the graph it can be seen that F = 2Fav ;
therefore, F = 120 N ;
Or
area under graph is "p = 0.71 N s ;
1
area is Fmax"t ;
2
2� 0.71
and so Fmax == 120 N ; [3]
12�10-3
  13  M06/4/PHYSI/SP2/ENG/TZ1/XX/M+
Part 2 Radioactive decay
(a) (i) isotopes of elements are chemically identical but have different atomic masses /
OWTTE / same number of protons in the nucleus but different number of
neutrons / OWTTE; [1]
(ii) the time it takes for the activity to halve / the time it takes for half the
number of radioactive atoms to decay / OWTTE; [1]
(iii) proton/1H / p+ ; [1]
1
(b) (i) no more C-14/carbon dioxide is taken in when a tree is dead;
the amount of C-14 determines the activity (of the charcoal);
since C-14 is radioactive the amount present (in the charcoal) decreases with
time / OWTTE; [3]
(ii) 14 g of C-14 contains 6.03�1023 atoms;
6.03�1023 �10-12
therefore, 10-12 contains H" 4�1010 ; [2]
14
Or
Mass of atom = 14�1.661�10-27 kg ;
So number of atoms in 1�10-12 g is
1�10-12 �10-3
;
14�1.661�10-27
= 4�1010
  14  M06/4/PHYSI/SP2/ENG/TZ1/XX/M+
(c) 4
3
number of
2
atoms�1010
1
0
0 0.40 0.80 1.2 1.6 2.0
time / years�104
�#The data points at 4.00 �1010 and
four data points covering 3 half-lives;
Ź#
0.50 �1010 must be shown.
�#
correct plotting of data points;
line of best fit to 1.8�104 years; [3]
1.9� 4�1010
(d) (i) number of atoms = H" 0.8�1010 ; [1]
9.6
(ii) from the graph age =1.3 (ą 0.2)�104 years; [1]
Allow ECF from (c) and from (d)(i).
  15  M06/4/PHYSI/SP2/ENG/TZ1/XX/M+
B3. Part 1 Electric motor
(a) (i) tension in thread
weight of object
tension in thread;
weight (of object) / mg;
tension length > weight length; [3]
2s
(ii) a = ;
t2
2� 0.84
== 0.35ms-2 ;
(2.2)2
T - mg = ma ;
T = m(g + a) = 0.015�10.35 = 0.16 N ; [4]
(b) (i) measure the time it takes the object to go successive distances of say 10 cm / any
realistic length given or implied;
if the times are equal then speed is constant / OWTTE; [2]
(ii) increase in potential energy = 0.015�10� 0.84 = 0.13J ;
0.13
rate of working = power input = = 0.037 W ; [2]
3.4
(iii) power input to motor = VI = 6.0� 0.045 = 0.27 W ;
Pout 0.037
Eff = = = 0.14 or 14% ; [2]
Pin 0.27
  16  M06/4/PHYSI/SP2/ENG/TZ1/XX/M+
Part 2 Waves
(a) the direction in which energy (of the wave) is propagated;
for a transverse wave it is at right angles to the direction of vibration of the
particles (of the medium through which the wave is travelling);
for a longitudinal wave the direction of energy propagation is in the same
direction as the vibration of the particles; [3]
Accept answers based on diagrams for full marks provided direction of energy
transfer and direction of oscillation are clear on the diagram.
(b) (i) longitudinal;
it is likely that the hammer will set the atoms of the rod to vibrate in the
same direction as the direction of the motion of the hammer / OWTTE; [2]
Award [0] if no explanation or poor explanation.
Or
hammer would not experience a rebounding force (if wave were not longitudinal)
/OWTTE;
some reference to direction of propagation of energy being along the length of the rod;
(ii) s = 3.00 m ;
s 3.00
v = = = 5.00�103 ms-1 ; [2]
t 6.00�10-4
Watch out for incorrect answers based on v = f  and
1
f ==1667 Hz ! It can give the correct numerical result with a
6�10-4
completely wrong argument.
(iii) the hammer blow/pulse sets the rod vibrating;
the vibration of the rod causes the air molecules in contact with the rod to
vibrate;
thereby setting up a longitudinal wave in the air / creates the sound /
OWTTE; [3]
(iv)  = 2l = 3.00 m ;
v 5.00�103
f = = =1.67�103 Hz; [2]
 3.00